SOLUTION: {{{2/(y^2-1)}}}+3+{{{1/(y+1)}}}

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: {{{2/(y^2-1)}}}+3+{{{1/(y+1)}}}      Log On


   

Question 178060: 2%2F%28y%5E2-1%29+3+1%2F%28y%2B1%29
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Assume you mean
2%2F%28y%5E2-1%29 + 3 + 1%2F%28y%2B1%29
:
(y^2-1) is the difference of squares, factor
2%2F%28%28y-1%29%28y%2B1%29%29 + 3 + 1%2F%28y%2B1%29
:
common denominator would be (y-1)(y+1)
%282+%2B+3%28y-1%29%28y%2B1%29+%2B+%28y-1%29%29%2F%28%28y%2B1%29%28y-1%29%29
:
%282+%2B+3%28y%5E2-1%29+%2B+%28y-1%29%29%2F%28%28y%2B1%29%28y-1%29%29
:
%282+%2B+3y%5E2+-+3+%2B+y+-+1%29%2F%28%28y%2B1%29%28y-1%29%29
:
combine like terms
%283y%5E2+%2B+y+-+2%29%2F%28%28y%2B1%29%28y-1%29%29
:
numerator will factor:
%28%283y-2%29%28y%2B1%29%29%2F%28%28y%2B1%29%28y-1%29%29
:
cancel (y+1)'s
%28%283y-2%29%29%2F%28%28y-1%29%29